Bioequivalence and Bioavailability Forum

Hi Loky do,

❝ ❝ ... it is considered acceptable that low-lier profiles can be excluded from statistical analysis […] if they occur with the same or lower frequency in the test product compared to the reference product.

❝

❝ Is this case-specific for this product, and its formula?

Yes. It’s based on numerous replicate design studies reviewed by the PKWP. AFAIK, in all more outliers were seen after the reference. It’s a terrible product showing also funky batch-to-batch variability.

❝ or it could be followed for other products with the same cases?

If it’s a delayed release product, see my previous post. However, note the subtle different wording “number” vs. “frequency”.

In the draft MR-GL “comparable frequency” was used as well. From a statistical point of view that calls for a two-sided test. When I showed an example at the GL-meeting in Bonn (2013) I stirred up a hornets’ nest. Imagine: 64 subjects, 4 outliers after T and 3 after R. Is the frequency comparable? Yes, the p-value is 0.7139!¹ The members of the PKWP did not like that at all. Oops, one (!) more. Changed in the final GL. Not acceptable. You can have it even more extreme. A 4-period full replicate study, one outlier (0.78%) after T and none after R → p-value 0.5.² Not acceptable because the bloody number is higher!
Lesson learned: Bad science (statistics is taboo). If you have one more outlier after T than after R – independent of the sample size – you’re dead.

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1. Two-sided test (is T = R?)
   subj <- 64
per  <- 2
seq  <- 2
OL.R <- 3
OL.T <- 4
n.T  <- n.R <- subj*per/seq
outliers  <- matrix(c(OL.T, n.T-OL.T, OL.R, n.R), nrow = 2,
                    dimnames=list(Guess = c("T", "R"),
                                  Truth = c("T", "R")))
fisher.test(outliers, alternative = "two.sided")

        Fisher's Exact Test for Count Data
data:  outliers
p-value = 0.7139
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
  0.2296229 10.0829801
sample estimates:
odds ratio
  1.418408
   
   
2. One-sided test (is T > R?)
   subj <- 64
per  <- 4
seq  <- 2
OL.R <- 0
OL.T <- OL.R + 1
n.T  <- n.R <- subj*per/seq
outliers  <- matrix(c(OL.T, n.T-OL.T, OL.R, n.R), nrow = 2,
                    dimnames=list(Guess = c("T", "R"),
                                  Truth = c("T", "R")))
fisher.test(outliers, alternative = "greater")

        Fisher's Exact Test for Count Data

data:  outliers
p-value = 0.5
alternative hypothesis: true odds ratio is greater than 1
95 percent confidence interval:
 0.05262625        Inf
sample estimates:
odds ratio
       Inf